AMS 311 (Fall, 2009)Joe MitchellPROBABILITY THEORYHomework Set # 2 – Solution Notes(1).(12 points)Suppose that two fair dice have been tossed and the total of their top faces is found to bedivisible by 4. What is the probability that both of them have landed 6?The sample space consists of 36 equally likely outcomes,S={(1,1),...,(6,6)}. We want to computeP(E|F), whereE= “both are 6”={(6,6)}andF= “sum is divisible by 4” ={(1,3), (2,2), (2,6), (3,1),(3,5), (4,4), (5,3), (6,2), (6,6)}.P(E|F) =P(E∩F)P(F)=P({(6,6)})P({(1,3),(2,2),(2,6),(3,1),(3,5),(4,4),(5,3),(6,2),(6,6)})=19(2).(12 points)Suppose for simplicity that the number of children in a family is 1, 2, or 3, with probability1/3 each. Little Bobby has no sisters. What is the probability that he is an only child? (Set the problem upcarefully. Remember to define the sample space, and any events that you use!)We can define the sample space asS={B, G, BB, BG, GB, GG, BBB, BBG, BGB, BGG, GBB, GBG,GGB, GGG}. LetA={B,BB,BBB}be the event that there are no girls in the family (ie, Bobby has nosisters). LetBibe the event that the family hasichildren. (e.g.,B1={B}andB2={BB,BG,GB,GG})P(B1∩A) =P(B1)·P(A|B1) = (1/3)(1/2) = 4/24P(B2∩A) =P(B2)·P(A|B2) = (1/3)(1/4) = 2/24P(B3∩A) =P(B3)·P(A|B3) = (1/3)(1/8) = 1/24We want to compute:P(B1|A) =P(B1∩A)P(A)=4/244/24 + 2/24 + 1/24= 4/7(3).(12 points)English and American spellings arecolourandcolor, respectively.A man staying at aParisian hotel writes this word, and a letter taken at random from his spelling is found to be a vowel.If40 percent of the English-speaking men at the hotel are English and 60 percent are Americans, what is theprobability that the writer is an Englishman?

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